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If the set of all solutions of $$|x^2 + x - 9| = |x| + |x^2 - 9|$$ is $$[\alpha, \beta] \cup [\gamma, \infty)$$, then $$(\alpha^2 + \beta^2 + \gamma^2)$$ is equal to :
The given equation is
$$|x^{2}+x-9| = |x| + |x^{2}-9|$$
The expressions inside the absolute values change sign at the real zeros of
$$x = 0,\quad x = \pm 3,\quad x = \frac{-1\pm\sqrt{37}}{2}\;(\approx -3.541,\; 2.541).$$
Hence analyse the equation in the six intervals determined by these points.
Here $$x<0,\;x^{2}-9>0,\;x^{2}+x-9>0$$ giving
$$x^{2}+x-9 = -x + (x^{2}-9)\implies x=0$$
No value in this interval satisfies the equation.
Case 2: $$\frac{-1-\sqrt{37}}{2}\lt x\lt -3.$$Now $$x<0,\;x^{2}-9>0,\;x^{2}+x-9<0$$ so
$${-(x^{2}+x-9)} = -x + (x^{2}-9)$$ $$\Longrightarrow 2x^{2}=18\Longrightarrow x=\pm3.$$
Neither value lies in this open interval, hence no solution here.
Case 3: $$-3\lt x\lt 0.$$Here $$x<0,\;x^{2}-9<0,\;x^{2}+x-9<0$$ giving
$${-(x^{2}+x-9)} = -x + (-(x^{2}-9))$$ $$\Longrightarrow -x^{2}-x+9 = -x^{2}-x+9,$$
which is an identity. Therefore every $$x$$ in $$(-3,0)$$ is a solution.
Case 4: $$0\lt x\lt\frac{-1+\sqrt{37}}{2}\;(\;0\lt x\lt 2.541\;).$$Now $$x>0,\;x^{2}-9<0,\;x^{2}+x-9<0$$ so
$${-(x^{2}+x-9)} = x + (-(x^{2}-9))$$ $$\Longrightarrow -x^{2}-x+9 = -x^{2}+x+9$$ $$\Longrightarrow x=0.$$
Again, no interior point meets the requirement; only the boundary point $$x=0$$ works.
Case 5: $$\frac{-1+\sqrt{37}}{2}\lt x\lt 3.$$Here $$x>0,\;x^{2}-9<0,\;x^{2}+x-9>0$$ so
$$x^{2}+x-9 = x + (-(x^{2}-9))$$ $$\Longrightarrow 2x^{2}=18\Longrightarrow x=\pm3.$$
The only admissible value would be $$x=3,$$ but $$x=3$$ is not inside this open interval.
Case 6: $$x\ge 3.$$Now all three quantities are non-negative, so
$$x^{2}+x-9 = x + (x^{2}-9),$$
which is always true. Hence every $$x\ge3$$ satisfies the equation.
Finally test the remaining critical points:
• $$x=-3:\;|(-3)^2-3-9| = 3,\;|{-3}|+|9-9| = 3\Rightarrow$$ true.
• $$x=0:\;|0-9| = 9,\;0+|{-9}| = 9\Rightarrow$$ true.
• $$x=3:\;|9+3-9| = 3,\;3+0 = 3\Rightarrow$$ true.
Therefore the complete solution set is
$$[-3,0]\cup[3,\infty).$$
Identifying $$\alpha=-3,\;\beta=0,\;\gamma=3,$$ we have
$$\alpha^{2}+\beta^{2}+\gamma^{2}=(-3)^{2}+0^{2}+3^{2}=9+0+9=18.$$
Option B which is: 18
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